connectednesscovering-spacesgeneral-topology
Show that if $p:X\rightarrow Y$ is a covering map and $Y$ is locally path-connected, then so is $X$.
How do you go about proving this? I can think of two ways of doing this, either by definition of locally path-connected and covering map or by trying to use the lift.
I've tried writing out the definitions of both concepts but even then, I don't see how I can "pull back" the path-connected set into the domain.
One counterexample is a variant on the famous topologist's sine curve.
Consider the graph of $y = \sin(\pi/x)$ for $0<x<1$, together with a closed arc from the point $(1,0)$ to $(0,0)$:
This space is obviously path-connected, but it is not locally path-connected (or even locally connected) at the point $(0,0)$.
Lemma: Given the commutative diagram
$$\begin{array}{ccccccccc} \widetilde{X} & \\ \downarrow{\small{p}} & {\searrow}^{q} \\ X_1 & \!\!\!\!\! \xleftarrow{p_1} & \!\!\!\! X_2\end{array}$$
where $p_1, p$ are covering maps, then so is $q$, where $X_1, X_2, \tilde{X}$ are all path-connected and locally path-connected.
Proof:
$q$ is surjective: $\sigma$ be a path in $X_2$ from $x_0$ and $x$. Pushforward by $p_1$ to get a path $p_1 \circ \sigma$ in $X_1$ from $p_1(x_0) = x_0'$ to $p_1(x)$. Lift to $\widetilde{X}$ to get a path $\widetilde{\sigma}$ starting from some point $x_0''$ in the fiber over $x_0'$. Pushforward by $q$ to get path $q \circ \widetilde{\sigma}$ starting at $x_0$. Uniqueness of path-lifing says $q \circ \widetilde{\sigma} \simeq \sigma$, so that $q$ maps the endpoint of $\tilde{\sigma}$ to the endpoint $x$ of $\sigma$. As $X_2$ is path connected, we can apply this argument for all $x \in X_2$ to prove $q$ is surjective.
$q$ is a covering map: Pick $x \in X_2$. Pushforward by $p_1$ to get $p_1(x)$ in $X_1$. There is a path-connected neighborhood $\mathscr{U}$ of $p_1(x)$ evenly covered by $p_1$ and $p$ (take neighborhoods evenly covered by $p_1$ and $p$ and take intersection). $\mathscr{V}$ be the slice in $p_1^{-1}(\mathscr{U})$ containing $x$. $\{\mathscr{U}_\alpha\}$ be the slices in $p^{-1}(\mathscr{U})$. $q$ maps each slice $\mathscr{U}_\alpha$ to distinct slices in ${p_{1}}^{-1}(\mathscr{U})$. $q^{-1}(\mathscr{V})$ is then union of slices in $\{\mathscr{U}_\alpha\}$ which are mapped homeomorphically onto $\mathscr{V}$. I claim all $\mathscr{U}_\alpha$ are mapped homeomorphically on $\mathscr{V}$ by $q$. This can be proved slicewise, recalling that given a commutative diagram with any two arrows as homeomorphism, so is the third. $\blacksquare$
If $\widetilde{X}$ is simply connected, $p : \widetilde{X}\to X$ the universal cover, $p_1 : X_2 \to X_1$ a covering map, then as $p_*(\pi_1(\widetilde{X}))$ fits inside ${p_1}_*(\pi_1(X_2))$, being the trivial group, we can lift $p$ to $\tilde{p} : \widetilde{X} \to X_1$. By previous discussion, $\tilde{p}$ is a covering map, since it fits inside a commutative diagram like above. Thus, $\widetilde{X}$ covers $X_2$, as desired.
Best Answer
Let $x\in X$ and $V$ be an open neighborhood of $x$. There is an evenly covered open neighborhood $U$ of $p(x)$. Say $x\in U_\alpha$ and $p_\alpha:U_\alpha\to U$ is the restriction of $p$ and a homeomorphism. Then $V\cap U_\alpha$ is homeomorphic to $p(V\cap U_\alpha)$. This is an open set containing $p(x)$, so it contains a path-connected neighborhood $W$. You can then just take the preimage of $W$ under $p_\alpha$.
Remark: The local (path-)connectedness of $X$ makes $Y$ local (path-)connected, too, which is very easy to prove since $p$ is continuous and open. But it can be shown that even a quotient map carries over the local (path-)connectedness from the domain to the image. (This is not difficult either, if you use the right characterization of local connectedness.)